$h(n) = -6n-1$ $g(x) = 7x^{2}+7x+7-5(h(x))$ $ g(h(1)) = {?} $
First, let's solve for the value of the inner function, $h(1)$ . Then we'll know what to plug into the outer function. $h(1) = (-6)(1)-1$ $h(1) = -7$ Now we know that $h(1) = -7$ . Let's solve for $g(h(1))$ , which is $g(-7)$ $g(-7) = 7(-7)^{2}+(7)(-7)+7-5(h(-7))$ To solve for the value of $g$ , we need to solve for the value of $h(-7)$ $h(-7) = (-6)(-7)-1$ $h(-7) = 41$ That means $g(-7) = 7(-7)^{2}+(7)(-7)+7+(-5)(41)$ $g(-7) = 96$